從一個表獲取兩個查詢的重疊

下面的PHP/MySQL的代碼工作，但它是笨重;兩個查詢，數據被轉換成數組，數組進行比較！

（請注意這是一個功能，因爲它返回是否有重疊）

$manager_sql = "SELECT league_id FROM league_members WHERE manager = 1 AND member_active = 1 AND player_id = $mi"; 
$player_sql = "SELECT league_id FROM league_members WHERE member_active = 1 AND player_id = $pi"; 

$manager_result = mysql_query($manager_sql) or die(mysql_error()); 
$manager_leagues = array(); 
while($sqlrow = mysql_fetch_array($manager_result)) { 
    extract($sqlrow); 
    $manager_leagues[] = $league_id; 
} 

$player_result = mysql_query($player_sql) or die(mysql_error()); 
$player_leagues = array(); 
while($sqlrow = mysql_fetch_array($player_result)) { 
    extract($sqlrow); 
    $player_leagues[] = $league_id; 
} 

$result = array_intersect($manager_leagues, $player_leagues); 

return count($result) > 0;

這裏是在一個去做查詢一些嘗試：

(SELECT league_id FROM league_members WHERE manager = 1 AND member_active = 2 AND player_id = '$mi') 
    UNION ALL 
(SELECT league_id FROM league_members WHERE member_active = 1 AND player_id = '$pi')

看起來工作，但返回太多行：

SELECT league_id FROM league_members WHERE league_id IN (
    SELECT league_id FROM league_members WHERE manager = 1 AND member_active = 1 AND player_id = '$mi' 
) AND league_id IN (
    SELECT league_id FROM league_members WHERE member_active = 1 AND player_id = '$pi' 
)

來源

2013-02-27 sq2

通過重疊，你的意思是什麼？ – 2013-02-27 05:59:08

如果你只是在第一個查詢中檢查'$ pi'，它會是同樣的事情。 – kennypu 2013-02-27 06:02:03

A $ mi管理着一堆聯賽（例如2,4,6），$ pi在一堆聯賽中打球（例如1,3,7） - 重疊比較2,4,6和1,3， 7 – sq2 2013-02-27 06:06:27

嘗試此查詢：

$sql = "SELECT ls1.league_id FROM league_members AS ls1 
     INNER JOIN league_members AS ls2 ON ls1.league_id = ls2.league_id 
     WHERE ls1.manager = 1 AND ls1.member_active = 1 AND ls1.player_id = '".$mi."' 
     AND ls2.member_active = 1 AND ls2.player_id = '".$pi."'"

注：

mysql_ *函數deprectated，使用mysqli_ *函數或PDO
您的查詢很容易受到SQL注入，使其seccure。

來源

2013-02-27 06:07:36

他的查詢不一定是脆弱的，它取決於$ mi和$ pi的數據來自哪裏。如果說，他們已經投到（int）s，那麼注射應該不成問題。但我同意這是不好的做法。 – Jason 2013-02-27 06:13:09

感謝您的所有意見。 $ mi和$ pi只能是整數，因此安全。（着名的遺言） – sq2 2013-02-27 06:16:06

另一方面，僅僅因爲你認爲'$ mi'和'$ pi'只能是整數並不能實現，你必須執行這個約束。我說的是，你的應用程序今天可能還行，但隨着時間的推移，你可能會將'$ mi'和'$ pi'的來源從現在改爲不太可靠的來源，如果你避難在代碼中明確強制實施整數約束，那麼當時你可能會忘記這麼做。 – Jason 2013-02-27 14:30:05

你希望找到任何league_id的同時具有$mi作爲經理和$pi作爲的球員之一？

SELECT DISTINCT l.league_id FROM league_members l JOIN league_members l2 ON l.league_id = l2.league_id AND l.member_active = 1 AND l2.member_active = 1 AND l.manager = 1 AND l2.manager = 0 WHERE l.player_id = $mi AND l2.player_id = $pi;

來源

2013-02-27 06:08:16 Jason

嘗試

"SELECT league_id FROM league_members WHERE manager = 1 AND member_active = 1 AND (player_id = '".$mi."' OR player_id = '".$pi."')";

來源

2013-02-27 06:08:47

請嘗試以下將它的工作。如果沒有，請讓我知道你有哪些問題。

SELECT league_id FROM league_members WHERE league_id IN（ SELECT league_id FROM league_members WHERE（member_active = 1 AND（player_id = '$ MI' 或player_id = '$ PI'）和管理者= 1）

來源

2013-02-27 06:09:20

我明顯沒有「T試過，但我建議最後查詢，與distinct關鍵字添加：

SELECT distinct league_id FROM league_members WHERE league_id IN (
    SELECT league_id FROM league_members WHERE manager = 1 AND member_active = 1 AND player_id = '$mi' 
) AND league_id IN (
    SELECT league_id FROM league_members WHERE member_active = 1 AND player_id = '$pi' 
)

來源

2013-02-27 06:11:23

這也適用 - 很高興知道我離得很近！ – sq2 2013-02-27 06:15:25

這個怎麼

SELECT league_id 
FROM league_members 
WHERE (manager = 1 AND member_active = 1 AND player_id = $mi) OR 
     (member_active = 1 AND player_id = $pi) 
GROUP BY league_id 
HAVING COUNT(*) = 2

來源

2013-02-27 06:18:36

此查詢是否有效？ – 2013-02-27 06:20:26

從一個表獲取兩個查詢的重疊

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